今天的題目是第3題,medium難度.

3. Longest Substring Without Repeating Characters

Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3.

Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

3. 無重復(fù)字符的最長子串

給定一個(gè)字符串，請(qǐng)你找出其中不含有重復(fù)字符的 最長子串 的長度。

示例 1:

輸入: "abcabcbb"
輸出: 3 
解釋: 因?yàn)闊o重復(fù)字符的最長子串是 "abc", 所以其長度為3。

示例 2:

輸入: "bbbbb"
輸出: 1
解釋: 因?yàn)闊o重復(fù)字符的最長子串是 "b"，所以其長度為1。

示例 3:

輸入: "pwwkew"
輸出: 3
解釋: 因?yàn)闊o重復(fù)字符的最長子串是 "wke"，所以其長度為3。請(qǐng)注意，你的答案必須是 子串 的長度，"pwke"是一個(gè)子序列，不是子串。

My answer:

首先對(duì)于查詢是否存在的操作我們選擇用dict來做(hash速度快), 對(duì)整個(gè)字符串進(jìn)行遍歷 用dict字典中存儲(chǔ)已經(jīng)訪問過的數(shù)據(jù). 對(duì)于未存在于dict中的元素直接添加key:value為s[i]:i; 當(dāng)遇到已經(jīng)存在的元素更新start的位置為dict[s[i]]的下一位, 因?yàn)閐ict中的值仍然保留start之前的數(shù)元素, 所以遇到的存在元素未必是有效的, 需要對(duì)start的更新值進(jìn)行判斷start = max(start, dct[s[i]] + 1). 最后更字典和最大長度即可.

class Solution:


    def lengthOfLongestSubstring(self, s: str) -> int:
        start = 0
        max_len = 0
        dct = {}
        for i in range(len(s)):
            if s[i] in dct:
                start = max(start, dct[s[i]] + 1) 
            dct[s[i]] = i
            max_len = max(max_len, i - start+1)
        return max_len